Problem: You are given an unweighted tree with n vertices. Recall that a tree is a connected undirected graph without cycles.

Your task is to choose three distinct vertices a, b, c on this tree such that the number of edges which belong to at least one of the simple paths between a and b, b and c, or a and c is the maximum possible. See the notes section for a better understanding.

The simple path is the path that visits each vertex at most once.

Input

The first line contains one integer number n (3 ≤ n ≤ 2 ⋅ 10^5) — the number of vertices in the tree. 

Next n - 1 lines describe the edges of the tree in form a_i, b_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i). It is guaranteed that given graph is a tree.

Output

In the first line print one integer res — the maximum number of edges which belong to at least one of the simple paths between a and b, b and c, or a and c.

In the second line print three integers a, b, c such that 1 ≤ a, b, c ≤ n and a ≠, b ≠ c, a ≠ c.

If there are several answers, you can print any.

Example

Input


8
1 2
2 3
3 4
4 5
4 6
3 7
3 8


Output


5
1 8 6

Note

The picture corresponding to the first example (and another one correct answer):

<image>

If you choose vertices 1, 5, 6 then the path between 1 and 5 consists of edges (1, 2), (2, 3), (3, 4), (4, 5), the path between 1 and 6 consists of edges (1, 2), (2, 3), (3, 4), (4, 6) and the path between 5 and 6 consists of edges (4, 5), (4, 6). The union of these paths is (1, 2), (2, 3), (3, 4), (4, 5), (4, 6) so the answer is 5. It can be shown that there is no better answer.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class F{
	
	static boolean[] vis;
	static int n;
	static List<Integer>[] adj;
	
	static int furthest;
	static int farNode;
	
	static void dfs1(int i, int p, int d) {
		if (d > furthest) {
			furthest = d;
			farNode = i;
		}
		for (int x: adj[i]) {
			if (x==p) continue;
			dfs1(x,i,d+1);
		}
	}
	
	static boolean dfs2(int i, int p, int t) {
		if (i == t) return vis[i] = true;
		for (int x: adj[i]) {
			if (x==p) continue;
			vis[i] |= dfs2(x,i,t);
		}
		return vis[i];
	}
	
	static void dfs3(int i, int p, int d) {
		if (d > furthest) {
			furthest = d;
			farNode = i;
		}
		for (int x: adj[i]) {
			if (vis[x] || x == p) continue;
			dfs3(x,i,d+1);
		}
	}
	
	public static void main(String[] args) throws IOException {
		// br = new BufferedReader(new FileReader(".in"));
		// out = new PrintWriter(new FileWriter(".out"));
		//new Thread(null, new (), "peepee", 1<<28).start();
		// Select diameter andthen longest path 
		n = readInt();
		adj = new List[n];
		for (int i = 0; i < n; i++) adj[i] = new ArrayList<Integer>();
		for (int i = 0; i < n-1; i++) {
			int x = readInt()-1;
			int y = readInt()-1;
			adj[x].add(y);
			adj[y].add(x);
		}
		farNode = furthest = 0;
		dfs1(0,0,0);
		List<Integer> b = new ArrayList<Integer>();
		b.add(farNode);
		furthest = 0;
		int cur = farNode;
		dfs1(farNode,farNode,0);
		b.add(farNode);
		long val = furthest;
		vis = new boolean[n];
		dfs2(farNode,farNode, cur);
		
		furthest = 0;
		for (int i =0 ; i < n; i++) {
			if (vis[i]) {
				//System.out.println("using this as jumpoff " + i + " " + adj[i]);
				dfs3(i,i,0);
			}
		}
		val += furthest;
		if (furthest== 0) {
			for (int i = 0; i < n; i++) if (!b.contains(i)) {
				farNode = i;
				break;
			}
		}
		b.add(farNode);

		out.println(val);
		
		for (int x: b) out.print(x +1+ " ");
		
		out.println();
		out.close();
	}
	
	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
	static StringTokenizer st = new StringTokenizer("");
	static String read() throws IOException{return st.hasMoreTokens() ? st.nextToken():(st = new StringTokenizer(br.readLine())).nextToken();}
	static int readInt() throws IOException{return Integer.parseInt(read());}
	static long readLong() throws IOException{return Long.parseLong(read());}
	static double readDouble() throws IOException{return Double.parseDouble(read());}
	
}